Q(x)=2x^2-4x-15

Solution for Q(x)=2x^2-4x-15 equation:

(Q)=2Q^2-4Q-15

We move all terms to the left:

(Q)-(2Q^2-4Q-15)=0

We get rid of parentheses

-2Q^2+Q+4Q+15=0

We add all the numbers together, and all the variables

-2Q^2+5Q+15=0

a = -2; b = 5; c = +15;
Δ = b2-4ac
Δ = 52-4·(-2)·15
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{145}}{2*-2}=\frac{-5-\sqrt{145}}{-4}
$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{145}}{2*-2}=\frac{-5+\sqrt{145}}{-4}
$